6 ms·

Does the law still work perfectly at very large distances considering single or zero photons (or gravitons or whatever) hitting a surface? I guess is there a m

• jakeinspace 7y ago
  Intuitively I'd think yes but only averaged over time. Waves in quantum fields should pretty much be perfectly matched to the inverse square law over any distance, but the actual observation of particles (by collapse or whichever interpretation you prefer) will obviously have variance which will only smooth out to an inverse square flux over time.
• evanb 7y ago
  The photons (let's say) that cause the inverse square law of the E+M force aren't on-shell photons [on-shell]. So, they're not photons that are "traveling" or are "emitted" in the same way that radiation is---they don't carry any energy away. Similar with or whatever. The 'force law' isn't susceptible to low-intensity counting issues. In the last case---radiation---the photons are on-shell (so you can, for example, count them) and inverse square does degrade when you hit low counts. However, time-averaging restores it on average. [on-shell] https://en.wikipedia.org/wiki/On_shell_and_off_shell https://en.wikipedia.org/wiki/On_shell_and_off_shell